Kan noen komme med løsningaforslag til disse Integralene:
t ln t dt
x/(x2+1) dt
integrert
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
[tex]\int t \ \cdot \ \ln{t} \rm{d}t[/tex]
Delvis integrasjon:
[tex]u^, = t \ , \ u = \frac{1}{2}t^2 \ , \ v = \ln{t} \ , \ v^, = \frac{1}{t}[/tex]
[tex]\int t \ \cdot \ \ln{t} \rm{d}t = \frac{1}{2} \ \cdot \ t^2 \ \cdot \ \ln{t} - \int \frac{1}{2}t^2 \ \cdot \ \frac{1}{t}\rm{d}t[/tex]
[tex]\int t \ \cdot \ \ln{t} \rm{d}t = \frac{1}{2} \ \cdot \ t^2 \ \cdot \ \ln{t} - \int \frac{1}{2}t\rm{d}t[/tex]
[tex]\int t \ \cdot \ \ln{t} \rm{d}t = \frac{1}{2} \ \cdot \ t^2 \ \cdot \ \ln{t} - \frac{1}{4}t^2 + C = \underline{\underline{\frac{1}{4}t^2(2\ln{t} - 1) + C}}[/tex]
[tex]\int \frac{x}{x^2 + 1} \rm{d}x[/tex]
Substitusjon:
[tex]u = x^2 + 1 \ , \ u^, = 2x \ \Rightarrow \ dx = \frac{du}{2x}[/tex]
[tex]\int \frac{\cancel{x}}{u} \ \cdot \ \frac{1}{2}\frac{\rm{d}u}{\cancel{x}}[/tex]
[tex]\frac{1}{2}\int \frac{1}{u} \rm{d}u = \frac{1}{2} \ \cdot \ \ln{|u|} + C[/tex]
[tex]\int \frac{x}{x^2 + 1}\rm{d}x = \underline{\underline{\frac{1}{2}\ln{(x^2 + 1)} + C}}[/tex]
Delvis integrasjon:
[tex]u^, = t \ , \ u = \frac{1}{2}t^2 \ , \ v = \ln{t} \ , \ v^, = \frac{1}{t}[/tex]
[tex]\int t \ \cdot \ \ln{t} \rm{d}t = \frac{1}{2} \ \cdot \ t^2 \ \cdot \ \ln{t} - \int \frac{1}{2}t^2 \ \cdot \ \frac{1}{t}\rm{d}t[/tex]
[tex]\int t \ \cdot \ \ln{t} \rm{d}t = \frac{1}{2} \ \cdot \ t^2 \ \cdot \ \ln{t} - \int \frac{1}{2}t\rm{d}t[/tex]
[tex]\int t \ \cdot \ \ln{t} \rm{d}t = \frac{1}{2} \ \cdot \ t^2 \ \cdot \ \ln{t} - \frac{1}{4}t^2 + C = \underline{\underline{\frac{1}{4}t^2(2\ln{t} - 1) + C}}[/tex]
[tex]\int \frac{x}{x^2 + 1} \rm{d}x[/tex]
Substitusjon:
[tex]u = x^2 + 1 \ , \ u^, = 2x \ \Rightarrow \ dx = \frac{du}{2x}[/tex]
[tex]\int \frac{\cancel{x}}{u} \ \cdot \ \frac{1}{2}\frac{\rm{d}u}{\cancel{x}}[/tex]
[tex]\frac{1}{2}\int \frac{1}{u} \rm{d}u = \frac{1}{2} \ \cdot \ \ln{|u|} + C[/tex]
[tex]\int \frac{x}{x^2 + 1}\rm{d}x = \underline{\underline{\frac{1}{2}\ln{(x^2 + 1)} + C}}[/tex]